Answer to Question #266341 in Differential Equations for Torjan

Question

Solve( D^2+ DD'-6D'^2)z= cos(2x+y)

Answer 1

$(D^2+DD'-6D'^2)z=cos(2x+y)$

The auxiliary equation is $m^2+m-6=0$

$m^2-2m+3m-6=0$

$m(m-2)+3(m-2)=0$

$(m+3)(m-2)=0$

Solved to get $m=2 , -3$

$\therefore$ the C.F is f₁(y+2x)+xf₂(y-3x)

P.I =$\frac{1}{D^2+DD'-6D'^2}cos(2x+y)$

Replacing D²=-a², DD'=-ab, D'²=-b²

From a=2, b=1

D²=-4, DD'=-2, D'²=-1

$P.I=\frac{1}{D^2+DD'-6D'^2}cos(2x+y)$

Incase of failure of f(D,D')=f(a,b)=0 for a=2, b=1 in cos(2x+y)

We therefore differentiate f(D,D') with respect to D and multiply f(x,y) by x

$P.I=\frac{1}{2D+D'}xcos(2x+y)$

Now replace D by a=2 and D' by b =1

$P.I=\frac{1}{2(2)+(1)}xcos(2x+y)$

$=\frac{x\ cos(2x+y)}{5}$

$z=C.F+P.I$

$\therefore z=f_1(y+2x)+xf_2(y-3x)+\frac{x\ cos(2x+y)}{5}$